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Standard XII

Chemistry

Law of Mass Action

Consider the ...

Question

Consider the following reaction equilibrium: $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ . Initially, 1 mole of $N_{2}$ and 3 moles of $H_{2}$ are taken in a 2 litre flask.

At equilibrium state, if the number of moles of $N_{2}$ is 0.6, then the total number moles of all gases present in the flask are:

0.8

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1.6

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2.8

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3.2

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Solution

The correct option is B 3.2
Out of 1 mol of nitrogen and 3 moles of hydrogen present initially, 0.4 moles of nitrogen and 1.2 moles of hydrogen react to form 0.8 moles of ammonia.

Thus, the number of moles at equilibrium is 0.6-mole nitrogen, 1.8-mole hydrogen, and 0.8 moles ammonia.

The total number of moles of all the gases present in the flask is $0.6 + 1.8 + 0.8 = 3.2$ .

Hence, the correct option is $D$

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Consider the following reaction equilibrium: N2(g)+3H2(g)⇌2NH3(g). Initially, 1 mole of N2 and 3 moles of H2 are taken in a 2 litre flask. At equilibrium state, if the number of moles of N2 is 0.6, then the total number moles of all gases present in the flask are: